kmp-最小子串回文次数

Posted 2020-10-09

poj 2406

Given two strings a and b we define a*b to be their concatenation. For example, if a = "abc" and b = "def" then a*b = "abcdef". If we think of concatenation as multiplication, exponentiation by a non-negative integer is defined in the normal way: a^0 = "" (the empty string) and a^(n+1) = a*(a^n).

Input

Each test case is a line of input representing s, a string of printable characters. The length of s will be at least 1 and will not exceed 1 million characters. A line containing a period follows the last test case.

Output

For each s you should print the largest n such that s = a^n for some string a.

Sample Input

abcd
aaaa
ababab
.

Sample Output

1
4
3

题目大意 ：
　　
　　此题就可以借助 kmp 的 next数组，因为 next 数组所表示的意义就是当前位置前面的串的最长公共前后缀，所以最后一位中的 next 所存的值就表示其前面的前缀的最长公共部分 是多少 ，用总的长度减去前缀的长度 ，就得出了最小回文的串的长度。

代码示例 ：

/*
 * Author:  ry 
 * Created Time:  2017/10/5 7:29:46
 * File Name: 1.cpp
 */
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <time.h>
#include <map>
using namespace std;
const int eps = 1e6+5;
#define ll long long

char s[eps];
int next[eps];
int len;

void getnext(){
    int i = 0, j = -1;
    next[0] = -1;
    while (i != len){
        if (j == -1 || s[i] == s[j]){
            i++;
            j++;
            next[i] = j;
        }
        else j = next[j];
    }
}

int main() {
    
    while (gets(s) != NULL){
        if (s[0] == ‘.‘) break;
        len = strlen(s);
        getnext();   
        int ff = len - next[len];
        if ((len - ff) % ff == 0) printf("%d\n", len/ff);
        else printf("1\n");
    }
    
    return 0;
}

/*
sadjkghrfkjashioawruiofhasjklryqwodhasnkbfgakjsrhoqwuiyeaskjbtrjksa   
*/